For the general quadractic equation,
[tex]ax^2+bx+c=0[/tex]
the roots are given by the quadratic formula,
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
In our case,
[tex]\begin{gathered} a=-10 \\ b=12 \\ c=-9 \end{gathered}[/tex]
Then, by substituting these values into the formula, we have
[tex]x=\frac{-12\pm\sqrt[]{12^2-4(-10)(-9)}}{2(-10)}[/tex]
which gives
[tex]\begin{gathered} x=\frac{-12\pm\sqrt[]{144^{}-360}}{-20} \\ x=\frac{-12\pm\sqrt[]{-216}}{-20} \\ x=\frac{6}{10}\pm\frac{6\sqrt[]{6}i}{-20} \end{gathered}[/tex]
